The size of that determinant isn’t the issue for inverting the matrix. What is an issue is what’s known as the conditioning of the matrix. Also you might want to use e.g. LinearSolve or a least-squares solver in preference to inverting.

You are setting precision higher than that of the input. This means some bits are effectively junk. As for assessing the condition number, one way is like so.

{vv, ss, ww} = SingularValueDecomposition[g];
sv = Diagonal[ss];
Length[sv]
sv[[1]]/sv[[-1]]

Out[16]= 100

Out[17]= 3.2296175965846435779*10^18

This tells you that computations will be effectively meaningless, insofar as they might lost 18 digits of precision when they only had around 16 to begin with,